Question

AD is the median of a triangle ABC and AE is perpendicular to BC.
Prove that AB ² = AD² – BC . DE + (BC/2)^2

[If AC >AB]​

Answer 1

Solution:-

D is the midpoint of BC and AE ⊥ BC.

In right angled triangle ABE,

AB² = AE² + BE² ....(1) - Pythagoras Theorem
AD² = AD² + ED² ....(2) - Pythagoras Theorem

From (1) and (2), we get

AB² = AD² - MD² + BE²

⇒ AB² = AD² - DE² + (BD - DE)²
⇒ AB² = AD² - DE² + BD² + DE² - 2BD × DE
⇒ AB² = AD² - 2BD × DE + BD²
⇒ AB² = AD² - 2(BC/2) × DE + (BC/2)²   {∵ BD = DC = BC/2}
⇒ AB² = AD² - BC × DE + BC²/4

Hence proved.

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